# How do you prove that if 2 vectors are perpendicular to each other, the absolute value of the sum equals the absolute value of the difference?

My answer is quite extensive.If you find it too technical at the beginning you have to read the last piece and skip the ‘ Prietpraat ‘.

The easiest is to do this if you use the concept of and and the standard derived therefrom.I will discuss how for vectors [MATH\\MATHBF x, y, z \\in \\mathbbR ^ n [/math.

If two vectors [math\\mathbf x, y [/math are perpendicular to each other, their and [math < \\mathbfx, \\mathbfy > [/math equals [Math0 [/math.

First the inproduct.An and [math <..,.. > [/math has the following properties, with [matha\\in \\mathbbr [/math a constant and [MATH\\MATHBF x, y, z \\In \\mathbbr ^ n [/math:

[Math < \\mathbfx, \\mathbfy > = < \\mathbfy, \\mathbfx > \\tag1 [/math

[Math < \\mathbfx + \\mathbfy, \\mathbfz > = < \\mathbfx, \\mathbfz > + < \\mathbfy, \\mathbfz > \\tag2 [/math

[Math < A\\MATHBF x, \\mathbfy > = a < \\mathbfx, \\mathbfy > \\tag * [/math

[Math < \\mathbfx, \\mathbfx > \\ge 0 \ \ Tag * [/math

And

[Math < \\mathbfx, \\mathbfx > = 0 \\Leftrightarrow \\mathbfx = \\mathbf0\\tag * [/math

So there is also:

[Math < \\mathbfz, \\mathbfx + \\mathbfy > = < \\mathbfz, \\mathbfx > + < \\mathbfz, \\mathbfy > \\tag * [/math

The last expression follows from (1) and (2).

As an example:

[MATH\\MATHBF x = \\beginbmatrixx_1\\\\x_2\\endbmatrix, \\mathbfy = \\beginbmatrixy_1\\\\y_2\\endbmatrix\\tag * [/math

We define as and (so it can be different, but this is what you initially learn if you have vectors in [math\\mathbb R ^ 2 [/math):

[Math < \\mathbfx, \\mathbfy > = x_1y_1 + x_2y_2\\tag * [/math

And

[Math < \\mathbfx, \\mathbfx > = x_1 ^ 2 + x_2 ^ 2 \ \ Tag * [/math

And we see that we really have an and here, because we meet the aforementioned requirements.(Check this).

You have learned that you can write down the length of a vector as follows:

[Math\\displaystyle | \\mathbfx | = \\sqrtx_1 ^ 2 + x_2 ^ 2 \\tag * [/math

Because we are squaring here (and taking a root), mathematicians call this a [mathL_2 [/math norm and actually note this differently:

[Math\\displaystyle \ \ | \\mathbfx\\ | _2 ^ 2 = x_1 ^ 2 + x_2 ^ 2 = < \\mathbfx, \\mathbfx > \\tag * [/math

The last expression reflects the relationship between the and and the norm (length) associated with this and:

[math\\ | \\mathbfx\\ | _2 ^ 2 = < \\mathbfx, \\mathbfx > \\tag * [/math

For vectors in [math\\mathbb R ^ n [/math , the Euclidean norm is often used, the shortest distance between 2 points, or as previously called The [mathL_2 [/math norm.

Thus, the length (squared) of the sum of 2 vectors can be found by:

[Math < \\mathbfx + \\mathbfy, \\mathbfx + \\mathbfy > = < \\mathbfx, \\mathbfx > + 2 < \\mathbfx, \\mathbfy > + < \\mathbfy, \\mathbfy > \\tag * [/math

Or if we fill in the norm:

[math\\ | \\mathbfx + \\mathbfy\\ | _2 ^ 2 = \ \ | \\mathbfx\\ | _2 ^ 2 + \ \ | \\mathbfy\\ | _2 ^ 2 + 2 < \\mathbfx, \\mathbfy > \\tag3 [/math

The length (squared) of the difference of 2 vectors:

[Math < \\mathbfx-\\mathbfy, \\mathbfx-\\mathbfy > = < \\mathbfx, \\mathbfx >-2 < \\mathbfx, \\mathbfy > + < \\mathbfy, \\mathbfy > \\tag * [/math

Or if we fill in the norm:

[math\\ | \\mathbfx-\\mathbfy\\ | _2 ^ 2 = \ \ | \\mathbfx\\ | _2 ^ 2 + \ \ | \\mathbfy\\ | _2 ^ 2-2 < \\mathbfx, \\mathbfy > \\tag4 [/math

You have to prove that expression (3) and (4) are equal, as you can see it is like this:

[Math < \\mathbfx, \\mathbfy > = 0 \ \ Tag * [/math

For vectors in [math\\mathbb R ^ n [/math applies:

[Math < \\mathbfx, \\mathbfy > = \ \ | \\mathbfx\\ | \ \ | \\mathbfy\\ | \\cos (\\theta) \\tag5 [/math

This means that this and is exactly [math0 [/math if the vectors each have an angle [Math\\theta [/math for which applies:

[Math\\cos (\\theta) = 0 \ \ Tag * [/math

If we assume that the vectors do not have the length (norm) 0.

And that is exactly when the vectors are perpendicular to each other.

Could this answer have been much shorter?To intuitively clarify what is going on, you only have to choose 2 mutually perpendicular vectors. The sum and the difference are the oblique side of an equally large triangle. The length of the oblique side of both triangles is therefore the same.

Try to sign this one!For example, choose:

[MATH\\MATHBF x = \\beginbmatrix1\\\\2\\endbmatrix, \\mathbfy = \\beginbmatrix2\\\\-1\\endbmatrix, \\mathbfx + y = \\beginbmatrix3\\\\1\\endbmatrix, \\mathbfx-y = \\beginbmatrix-1\\\\3\\endbmatrix\\tag * [/math

Check also that the and is 0.

So most of the above is actually Pythagoras ‘ theorem in extensive form.And the worst thing is that I have the most important step (5) not proof.

In contrast, I have put it in a broader context.A length (norm) can therefore be linked to an inproduct. Many of the above properties also apply to very different inproducts. Vectors can also look very different, they don’t have to be ‘ risers ‘ with numbers.